What is the product of two independent student t distributions? In which case does this product product result in another t distribution?

  • $\begingroup$ Because exact formulas are messy and there is no product that results in a t distribution, would you mind explaining the statistical motivation for this question? We might be able to suggest alternative approaches or more useful related questions. $\endgroup$
    – whuber
    May 27 at 13:41
  • $\begingroup$ Explaining why I need this product would be a little bit complicated, but I am looking for a special case when the product of 2 t distribution will give a t distribution or an unnormalized t distribution $\endgroup$
    – sam
    May 27 at 15:18
  • $\begingroup$ Could you clarify what an "unnormalized" t distribution might be? $\endgroup$
    – whuber
    May 27 at 15:21
  • $\begingroup$ I reformulated my problem so it will result in answering the following question: stats.stackexchange.com/questions/576853/… $\endgroup$
    – sam
    May 27 at 16:37
  • $\begingroup$ The distribution corresponding to the product of several $t$ densities is called a poly-t distribution. $\endgroup$
    – Xi'an
    May 28 at 6:55

1 Answer 1


When $X$ and $Y$ are independent random variables with densities $f_X$ and $f_Y,$ the density of their product can be found with a change of variables as

$$f_{XY}(z) = \int_{\mathbb R} f_X(x) f_Y(z/x)\,\frac{\mathrm{d}x}{|x|}.$$

Ignoring normalizing constants (we'll consider these at the end), for two Student t densities with $\nu$ and $\mu$ degrees of freedom this integrand is proportional to

$$h(x,z) = \left(1 + \frac{x^2}{\mu}\right)^{-(\mu+1)/2}\, \left(1 + \frac{z^2}{x^2\nu}\right)^{-(\nu+1)/2}\,\frac{1}{|x|}.$$

Let's find a lower bound for $f(z)$ when $z$ is small. To do so, we may restrict the region of integration and we may replace the integrand by anything that never exceeds it.

Let $z$ be positive but less than $1$ and consider the integration region where $x^2\nu$ ranges between $z^2$ and $1.$ To get an appreciation for what's going on, here (with $\mu=\nu=1$) are plots of $h(x,z)$ for $|z| = 1$ (blue), $1/2, 1/4,$ and $1/8$ (red).


You can see that as $|z|$ approaches $0,$ there's more and more area pushed into this region. That's no surprise: we would expect the largest area (which corresponds to the highest density of the product) to be at the center of the product distribution, which (by symmetry) must be $0.$ But how large does it get?

Over the region $x^2/\nu \in[z^2, 1]$ the first factor of $h$ is smallest when $x$ is smallest and the second factor is smallest when $x$ is largest, whence throughout this region

$$\begin{aligned} h(x,z) &\ge \left(1 + \frac{z^2}{\mu\nu}\right)^{-(\mu+1)/2}\, \left(1 + \frac{z^2}{1}\right)^{-(\nu+1)/2}\,\frac{1}{|x|} \\ &\ge \left(1 + \frac{1}{\mu\nu}\right)^{-(\mu+1)/2}\, \left(1 + 1\right)^{-(\nu+1)/2}\,\frac{1}{|x|}. \end{aligned}$$

The second inequality is a consequence of $z^2 \le 1.$

The factors before $1/|x|$ are constant (but nonzero), depending only on $\mu$ and $\nu,$ so again let's consider them later and ignore them now. As $x$ varies over just the positive part of this region it runs from $z/\sqrt{\nu}$ to $1/\sqrt{\nu},$ giving a lower bound proportional to

$$\int_{z/\sqrt{\nu}}^{1/\sqrt{\nu}} \frac{\mathrm{d}x}{|x|} = -\log z.$$

As $z\to 0,$ this lower bound diverges. Consequently, no matter what the constants of proportionality are that we ignored, $f_{XY}(z)$ diverges at $0.$

Here, to illustrate, is a histogram from a simulation of ten million products (with $\nu=\mu=1/2$). Almost a million of those products are represented. The red curve is the negative logarithm. Clearly it approximates the density well near zero.


However, any Student t distribution with (say) $\kappa \gt 0$ degrees of freedom has a value proportional to $(1 + 0^2/\kappa)^{-(\kappa+1)/2} = 1$ at the origin, which is finite. Consequently, the product of two independent Student t distributions is never (even remotely like) a Student t distribution.

The product density can be found analytically as a polynomial combination of Riemann hypergeometric functions. Since this product is never a Student t distribution, though, I did not see any point into providing further details.

  • $\begingroup$ Thank you for this detailed analysis. this makes sense now. For the special case when the degree of freedom is 1 therefore the t distribution is a cauchy distribution, it is possible to find the result of the product of two t(1)? $\endgroup$
    – sam
    May 27 at 17:38
  • $\begingroup$ @sam It seems so, see here: math.stackexchange.com/a/1451744 $\endgroup$ May 27 at 17:47
  • $\begingroup$ Yes, that result is correct. But generally, for arbitrary degrees of freedom, the PDF of the product does not attain such a simple form: various versions of hypergeometric functions show up, such as elliptic functions. $\endgroup$
    – whuber
    May 27 at 18:39

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